This element had a 3p sub level that contains 3 electrons

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311

LuckyWell [14K]

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

5 0

4 years ago

Balance the following redox reaction if it occurs in H2SO4. What are the coefficients in front of C3H8O2 and H2SO4 in the balanc

Mrac [35]

Answer:

3C3H8O2 and 16H2SO4

Explanation:

C3H8O2(aq) + K2Cr2O7(aq) → C3H4O4(aq) + Cr2(SO4)3(aq)

In acidic reaction you need to separate the half-reactions knows.

C3H8O2(aq) → C3H4O4(aq)

Cr2O7(2-) → Cr(+3) (aq)

Balance elements other than O and H. Add H2O to balance oxygen, balance hydrogen by adding protons (H+) and Balance the charge of each equation with electrons

C3H8O2(aq) + 2H2O(l) → C3H4O4(aq) + 8H(+) + 8e(-)

Cr2O7(2-) + 14H(+) + 6e(-) → 2Cr(+3) + 7H2O (aq)

Scale the reactions so that the electrons are equal. If not look for a common multiple. In this case is 24, 8*3=6*4. So for each equation multiply for the correct multiple. Then

3C3H8O2(aq) + 6H2O(l) → 3C3H4O4(aq) + 24H(+) + 24e(-)

4Cr2O7(2-) + 56H(+) + 24e(-) → 8Cr(+3) + 28H2O (aq)

Add the reactions and cancel out common terms and group the ions in compounds if u can.

3C3H8O2(aq) + 32H(+) + 4Cr2O7(2-) → 3C3H4O4(aq) + 4Cr2(SO4)3(aq) + 22H2O (aq)

We are in acidic medium so the H(+) must be from H2SO4. So 32H(+) is 16H2SO4 in reactives. Add SO4(2-) in products and balance

16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + 4SO4(2-)

Then

16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + + 2K2SO4 (aq)

8 0

4 years ago

Where does baby came from

Oksanka [162]

Answer:

they came from for the dad sperm

6 0

3 years ago

Imagine a ringingbell sat inside a sedet sans ja. Once all the arisrenoves and a vecums

kotykmax [81]

Answer:

what language is this is this english?

4 0

3 years ago

Summary of the periodic table

JulsSmile [24]

The periodic table to chart which arranges all of the unknown elements to the automatic numbers the columns in the periodic table are called families or groups the rows are called periods

4 0

3 years ago